# Homework #1 due Monday at 6pm. White drop box in Student Lounge on the second floor of Cory. Tuesday labs cancelled next week

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1 Announcements Homework #1 due Mondy t 6pm White drop ox in Student Lounge on the second floor of Cory Tuesdy ls cncelled next week Attend your other l slot Books on reserve in Bechtel Hmley, 2 nd nd 3 rd Edition EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 1 Review from Lst Clss KCL, KVL Node nd Loops Resistors in Series nd prllel Equivlent Resistnce Voltge nd Current Division I-V Chrcteristics Sources Resistors EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 2 1

2 Circuit w/ Dependent Source Exmple Find i 2, i 1 nd i o EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 3 Lecture #3 OUTLINE KCL, KVL Exmples Thevenin/Norton Equivlent circuit Mesurement Devices Reding Finish Chpter 1,2 EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 4 2

3 Using Equivlent Resistnces Simplify circuit efore pplying KCL nd/or KVL: Exmple: Find I 7 V - I R 1 R 2 R 4 R 5 R 3 R 6 R 1 = R 2 = 3 kw R 3 = 6 kw R 4 = R 5 = 5 kw R 6 = 10 kw EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 5 Node-Voltge Circuit Anlysis Method 1. Choose reference node ( ground ) Look for the one with the most connections! 2. Define unknown node voltges those which re not fixed y voltge sources 3. Write KCL t ech unknown node, expressing current in terms of the node voltges (using the I-V reltionships of rnch elements) Specil cses: floting voltge sources 4. Solve the set of independent equtions N equtions for N unknown node voltges EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 6 3

4 Nodl Anlysis: Exmple #1 R 1 R 3 - I S V 1 R 2 R 4 1. Choose reference node. 2. Define the node voltges (except reference node nd the one set y the voltge source). 3. Apply KCL t the nodes with unknown voltge. 4. Solve for unknown node voltges. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 7 Nodl Anlysis: Exmple #2 R 1 R 3 I 1 V R 5 V 1 R 2 R 4 V 2 EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 8 4

5 Nodl Anlysis w/ Floting Voltge Source A floting voltge source is one for which neither side is connected to the reference node, e.g. V LL in the circuit elow: V VLL - V I1 R2 R4 I2 EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 9 Nodl Anlysis w/ Floting Voltge Source Prolem: We cnnot write KCL t nodes or ecuse there is no wy to express the current through the voltge source in terms of V-V. Solution: Define supernode tht chunk of the circuit contining nodes nd. Express KCL for this supernode. Incorporte voltge source constrint into KCL eqution. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 10 5

6 Nodl Anlysis: Exmple #3 supernode V VLL - V I 1 R2 R 4 I 2 Eq n 1: KCL t supernode Sustitute property of voltge source: EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 11 Node-Voltge Method nd Dependent Sources If circuit contins dependent sources, wht to do? Exmple: i D 20 W 10 W 2.4 A 200 W 5i D 80 V EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 12 6

7 Node-Voltge Method nd Dependent Sources Dependent current source: tret s independent current source in orgnizing nd writing node eqns, ut include (sustitute) constrining dependency in terms of defined node voltges. Dependent voltge source: tret s independent voltge source in orgnizing nd writing node eqns, ut include (sustitute) constrining dependency in terms of defined node voltges. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 13 Exmple: i D 2.4 A 200 W 20 W 10 W 5i D 80 V EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 14 7

8 Mesh Circuit Anlysis Method 1) Select M independent mesh currents such tht t lest one mesh current psses through ech rnch* M = #rnches - #nodes 1 2) Apply KVL to ech mesh, expressing voltges in terms of mesh currents => M equtions for M unknown mesh currents 3) Solve for mesh currents => determine node voltges EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 15 Mesh Anlysis: Exmple #1 1. Select M mesh currents. 2. Apply KVL to ech mesh. 3. Solve for mesh currents. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 16 8

9 Mesh Anlysis with Current Source i i Prolem: We cnnot write KVL for meshes nd ecuse there is no wy to express the voltge drop cross the current source in terms of the mesh currents. Solution: Define supermesh mesh which voids the rnch contining the current source. Apply KVL for this supermesh. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 17 Mesh Anlysis: Exmple #2 i i Eq n 1: KVL for supermesh Eq n 2: Constrint due to current source: EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 18 9

10 Mesh Anlysis with Dependent Sources Exctly nlogous to Node Anlysis Dependent Voltge Source: (1) Formulte nd write KVL mesh eqns. (2) Include nd express dependency constrint in terms of mesh currents Dependent Current Source: (1) Use supermesh. (2) Include nd express dependency constrint in terms of mesh currents EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 19 Forml Circuit Anlysis Methods NODAL ANALYSIS ( Node-Voltge Method ) 1) Choose reference node 2) Define unknown node voltges 3) Apply KCL to ech unknown node, expressing current in terms of the node voltges => N equtions for N unknown node voltges 4) Solve for node voltges => determine rnch currents MESH ANALYSIS ( Mesh-Current Method ) 1) Select M independent mesh currents such tht t lest one mesh current psses through ech rnch* M = #rnches - #nodes 1 2) Apply KVL to ech mesh, expressing voltges in terms of mesh currents => M equtions for M unknown mesh currents 3) Solve for mesh currents => determine node voltges EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 20 10

11 Superposition A liner circuit is one constructed only of liner elements (liner resistors, nd liner cpcitors nd inductors, liner dependent sources) nd independent sources. Liner mens I-V chrcteristic of elements/sources re stright lines when plotted Principle of Superposition: In ny liner circuit contining multiple independent sources, the current or voltge t ny point in the network my e clculted s the lgeric sum of the individul contriutions of ech source cting lone. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 21 Superposition Procedure: 1. Determine contriution due to one independent source Set ll other sources to 0: Replce independent voltge source y short circuit, independent current source y open circuit Do not turn independent sources off!!! 2. Repet for ech independent source 3. Sum individul contriutions to otin desired voltge or current EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 22 11

12 Superposition Exmple Find V o 2 W 4 V 24 V 4 A 4 W V o EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 23 Equivlent Circuit Concept A network of voltge sources, current sources, nd resistors cn e replced y n equivlent circuit which hs identicl terminl properties (I-V chrcteristics) without ffecting the opertion of the rest of the circuit. network A of sources nd resistors i A i B v A v B i A (v A ) = i B (v B ) network B of sources nd resistors EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 24 12

13 Source Comintions Voltge sources in series cn e replced y n equivlent voltge source: v 1 v 2 v 1 v 2 Current sources in prllel cn e replced y n equivlent current source: i 1 i 2 i 1 i 2 EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 25 Thévenin Equivlent Circuit Any* liner 2-terminl (1-port) network of indep. voltge sources, indep. current sources, nd liner resistors cn e replced y n equivlent circuit consisting of n independent voltge source in series with resistor without ffecting the opertion of the rest of the circuit. Thévenin equivlent circuit R Th network of sources nd resistors v L i L R L V Th v L i L R L lod resistor EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 26 13

14 I-V Chrcteristic of Thévenin Equivlent The I-V chrcteristic for the series comintion of elements is otined y dding their voltge drops: For given current i, the voltge drop v i is equl to the sum of the voltges dropped cross the source (V Th ) nd cross the resistor (ir Th ) V Th R Th i v v = V Th ir v I-V chrcteristic of resistor: v = ir I-V chrcteristic of voltge source: v = V Th EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 27 Procedure to Find V Th 1. Clculte the open circuit voltge v OC from the originl circuit. v OC = V Th. Originl circuit Thévenin equivlent circuit R Th network of sources nd resistors v OC V Th v Th EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 28 14

15 Procedure to Find R Th 1. Turn off ll independent sources in originl circuit. Dependent source should e left s is. 2. Apply test voltge v test source etween terminls nd. Clculte the resulting current i test. 3. R th = v test / i test. Originl circuit network of sources nd resistors v test i test Thévenin equivlent circuit R Th v test i test EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 29 Thévenin Equivlent Voltge Exmple Find the Thevenin equivlent voltge with respect to the terminls,: EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 30 15

16 More wys of clculting R Th 1. Clculte I SC, the short circuit current flowing from to. R Th = V OC / I SC 2. Turn off ll independent sources nd find the equivlent resistnce R eq etween nd. R Th = R eq Network of Resistors And Sources (indep. Sources Enled) Method 1 I SC Network of Resistors (indep. Sources Disled) Method 2 R eq = R Th EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 31 Cution when clculting R Th using lgorithms from previous slide Method 1 is very quick, ut cnnot e used when I SC or V Th = 0 Method 2 cnnot e used when there re dependent sources in the network or when the resistive network is too complicted: R 1 R 3 R 5 R 2 R 4 Advise: Use V test nd I test method!!! EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 32 16

17 Comments on Dependent Sources A dependent source estlishes voltge or current whose vlue depends on the vlue of voltge or current t specified loction in the circuit. (device model, used to model ehvior of trnsistors & mplifiers) To specify dependent source, we must identify: 1. the controlling voltge or current (must e clculted, in generl) 2. the reltionship etween the controlling voltge or current nd the supplied voltge or current 3. the reference direction for the supplied voltge or current The reltionship etween the dependent source nd its reference cnnot e roken! Dependent sources cnnot e turned off for vrious purposes (e.g. to find the Thévenin resistnce, or in nlysis using Superposition). EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 33 Thevenin Equivlent Exmple #2 Find the Thevenin equivlent with respect to the terminls,: Methods 1 nd 2 fil in this cse. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 34 17

18 Networks Contining Time-Vrying Sources Cre must e tken in summing time-vrying sources! Exmple: 10 sin (100t) 1 kw 20 cos (100t) 1 kw EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 35 Norton Equivlent Circuit Any* liner 2-terminl (1-port) network of indep. voltge sources, indep. current sources, nd liner resistors cn e replced y n equivlent circuit consisting of n independent current source in prllel with resistor without ffecting the opertion of the rest of the circuit. Norton equivlent circuit network of sources nd resistors v L i L R L i N R N v L i L R L EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 36 18

19 I-V Chrcteristic of Norton Equivlent The I-V chrcteristic for the prllel comintion of elements is otined y dding their currents: For given voltge v, the current i is equl to the sum of the currents in i ech of the two rnches: i N R N i v i = -I N Gv v I-V chrcteristic of resistor: i=gv I-V chrcteristic of current source: i = -I N EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 37 Finding I N nd R N = R Th Anlogous to clcultion of Thevenin Eq. Ckt: 1) Find s.c. current nd Norton (Thev) resistnce Find I SC nd then use the V test / I test method to find R N 2) or find o.c voltge nd s.c. current (with cution!!) I N I sc = V Th /R Th EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 38 19

20 Finding I N nd R N We cn derive the Norton equivlent circuit from Thévenin equivlent circuit simply y mking source trnsformtion: R Th i L i L v Th v L R L i N R N v L R L v v R N = RTh = = i oc Th ; in = isc sc RTh EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 39 Mximum Power Trnsfer Theorem Thévenin equivlent circuit R Th Power sored y lod resistor: V Th v L i L R L p = i 2 L R L æ VTh = ç è RTh R L ö ø 2 R L dp To find the vlue of R L for which p is mximum, set to 0: dr dp ù L = V ú = 0 dr L û Þ Þ 2 é 2 ( RTh RL ) - RL 2( RTh RL ) Th ê 4 ë ( RTh RL ) 2 ( R R ) - R 2( R R ) = 0 R Th Th = R L L L Th L A resistive lod receives mximum power from circuit if the lod resistnce equls the Thévenin resistnce of the circuit. EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 40 20

21 Mesuring Voltge To mesure the voltge drop cross n element in rel circuit, insert voltmeter (digitl multimeter in voltge mode) in prllel with the element. Voltmeters re chrcterized y their voltmeter input resistnce (R in ). Idelly, this should e very high (typicl vlue 10 MW) Idel Voltmeter R in EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 41 Effect of Voltmeter undistured circuit circuit with voltmeter inserted V SS _ R 1 R 2 V V SS _ 2 V 2 ' R 1 R 2 R in V2 é R ù = V 2 SS ê ú ër1 R2 û V2 é R ù = 2 R V in SS ê ú ër2 Rin R1 û Exmple: VSS = 10 V, R2 = 100K, R1 = 900K Þ V2 = R = 1M 0 V =, i n 2 1V EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 42 21

22 Mesuring Current To mesure the current flowing through n element in rel circuit, insert n mmeter (digitl multimeter in current mode) in series with the element. Ammeters re chrcterized y their mmeter input resistnce (R in ). Idelly, this should e very low (typicl vlue 1W). Idel Ammeter R in EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 43 Effect of Ammeter V 1 Mesurement error due to non-zero input resistnce: undistured circuit _ R 1 I R 2 circuit with mmeter inserted V 1 _ R 1 I mes R in R 2 mmeter V1 V I = I 1 mes = R1 R2 R1 R2 Rin Exmple: V 1 = 1 V, R 1 = R 2 = 500 W, R in = 1W 1V I = m = A 1, I= mes 5 0W W0 EE40 Summer 2006: Lecture 3 Instructor: Octvin Florescu 44 22

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